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57:019:AAA Mechanics of Deformable Bodies Review Material

Overview: A. Key Ideas of Stress

B. Key Ideas of Strain

C. Hooke’s Law for Linear Isotropic Elastic Solids

D. Axial Loads and Deformation

E. Torsion and Twist

F. Bending Behaviors

1. V, M diagrams

2. Flexure Formulae

3. Composite Sections

4. Shear Stress & Shear Flow

5. Deflections

G. Euler Buckling

H. Energy Methods

I. Stress Transformations

A. Key Ideas of Stress

Stress is defined as force per unit area. When working with stresses, we need to be very specific

about the directionality of the forces and the directionality of the area on which the forces are

acting. This is done by representing stress as a rank-2 tensor as follows:

zzyzx

yzyyx

xzxyx

zzzyzx

yzyyyx

xzxyxx

τ

The diagonal entries in the stress tensor are said to be normal stresses because the associated force

component acts normal to the plane.

The off-diagonal entries in the stress tensor are called

shear stresses, because the associated force

components act parallel to the plane.

It is very straightforward to demonstrate that the stress

tensor is symmetric, or specifically that:

yxxy

zyyz

x y

z zz

xx

xz

xy

zy zx

yy

yz

yx

57:019 Review for Mechanics of Deformable Bodies Spring 2014

2

xzzx

B. Key Ideas of Strain 1. Normal strains Normal strain measures the change in length of an infinitesimal “fiber” of material relative to the

original length of that fiber.

0

0

ds

dsds

2. Shear Strains

Shear strain represents the negative change in angle (radians) during deformation between two

infinitesimal fibers that were initially perpendicular.

To describe the change in length of fibers aligned with Cartesian reference axes, and to quantify

change in angles between these fibers, a second rank strain tensor is used:

zzzyzx

yzyyyx

xzxyxx

ε

20

n

t

n’

’

t’

2

:strainshear nt

undeformed deformed

ds0

ds

undeformed fiber

deformed fiber

length of increase 0

length of change no 0

length ofreduction 0

;

;

;

; ; ;

21

21

21

21

21

21

21

21

21

y

u

z

u

x

u

z

u

x

u

y

u

z

u

y

u

x

u

zy

zyyzzyyz

zxzxxzzxxz

yxyxxyyxxy

zzz

y

yyx

xx


3

C. Hooke’s Law for Isotropic, Elastic Material

zx

yz

xy

zz

yy

xx

zx

yz

xy

zz

yy

xx

E

2

100000

02

10000

002

1000

00021

1

2121

0002121

1

21

000212121

1

1

zx

yz

xy

zz

yy

xx

zx

yz

xy

zz

yy

xx

E

)1(200000

0)1(20000

00)1(2000

0001

0001

0001

1

Example: The principal stresses at a point are shown in the figure. If the material is aluminum for which Eal=1*104 ksi and

al=0.33, determine the principal strains.

00277.26151010

1

000967.26151010

1

00237.26151010

1

26 ;15 ;10

31

31

4

31

31

4

31

31

4

zy

ksi

ksi

ksi

ksiksiksi

zz

yy

xx

x

15 ksi 10 ksi

26 ksi


4

D. Axial Strains and Deformations

Indeterminate Axially Loaded Members

Example: The bronze 86100 pipe

has an inner radius of 0.5 in. and a

wall thickness of 0.2 in. If the gas

flowing through it changes the

temperature of the pipe uniformly

from TA=200°F at A to TB=60°F at

B, determine the axial force it exerts

on the walls. The pipe was fitted

between the walls when T=60°F.

E=15,000ksi and 69.6 10 / F

LAB LBC

A B C P

x

P

A C FA FC

CAx FFPF 0

F

F AE

FLL

L

L

AE

F

E

AF

axialaxial

axial

0

0

0 0

0 0

( )( )

( )

( ) ( )( )

( ) ( ) ( )

( )( )

( ) ( )

L L

F xx

A x

x F xx

E x A x E x

F xL x dx dx

A x E x

x

Variable force, area Constant force, area

In this illustrative example, there is

only one relevant equation of

equilibrium 0 xF but there are

two support reactions at A and C to

solve for. Thus the system is statically

indeterminate.

To solve for both support reactions, we

need an additional condition. Here,

since the supports at both A and C are

taken to be rigid, we can say that

C/A=0 or that overall the axial member

does not change its length.


5

Solution:

From a free-body diagram, there are two equal and compressive wall forces FA=FB acting on the

pipe that tend to compress it. Alternatively, the temperature increase in the pipe generates

thermal strains that causes expansion. The resulting axial strain is as follows:

( ) ( ) BFx T x

AE where ( ) 1o

xT x T

L

and 140oT F

Because the two walls are rigid can be no change in length of the pipe.

0 0

( ) 0 1

L L

Bo

FxL x dx T dx

L AE

Solving for FB gives: 7.62

oB A

T AEF kip F

E. Torsion and Twist

axis centroidal smember'

thealong variablecoordinate theis x

member in the twist of angle theis

section-crosscircular theof radius theis c

axis centroidal thefrom

point material a of distance theis

strainshear smaterial' theis :where

max

cdx

d


6

Integrating the moment of shear stresses over a circular cross-section about the longitudinal axial of the

member we obtain the relation between shear stresses and torque:

Sign Convention:

Internal torques and their associated angle change are positive when the resultant vector points away from the cut face on which it acts.

L

GxJ

dxxT

0)(

)(JG

TL

2

0

4

max

2

2

A

c

T dA

d

c

c

max

GJ

T

dx

d

J

T

J

Tc

2

2

4

0

3

2

c

d

dAJ

c

A


7

Indeterminate Torsion Example:

F. Bending Behaviors

Shear and Bending Moment Diagrams

Bending occurs in members that experience loads perpendicular to the longitudinal axis.

( ) ( )

B

AB

A

dVw x V w x dx

dx

B

A

AB dxxVMxVdx

dM)()(

The flexure formula:

One-way bending: I

Mc

I

My

max ;

Biaxial Bending: y

y

z

z

I

zM

I

yMzy ),(

TA

TB

TC

From statics: TA+TB=TC

Additional constraint: AB=0

AB

ACCB

AB

BCCA

BCCCBACA

BCACACA

B

C

CB

C

A

AC

AB

L

LTT

L

LTT

LTLLT

JG

LTT

JG

LT

dxJG

Tdx

JG

T

;

0


8

Composite sections: Beams are often made of different materials in order to efficiently carry a load.

What is the elastic bending stress distribution on the composite cross-section? The cross section of the beam must be transformed into a single material if the flexure formula (which is based on homogenous materials) is to be used to compute the bending stress. A transformation factor “n=E1/E2” is used for this purpose. Once the section is transformed, I* is computed.

To calculate stresses on the cross-section:

In the original material: *

M y

I

. In the transformed material:

*

n M y

I

The transverse shear formula:

Shear Flow:

Shear Flow is denoted by “q” and denotes the shear force per unit length transmitted

along a specified longitudinal section.

Usually, we are interested in the shear flow along sections where different members

are joined.

Here, Q is the moment about the NA of the cross-sectional area connected to the

remainder of the section along the longitudinal section of interest.

A'

dA'y Q whereIt

VQτ

I

QVq

*


9

Example: The cantilever beam is subjected to the

loading shown, where P=7kN. Determine the average

shear stress developed in the nails within region AB of

the beam. The nails are located on each side of the beam

and are spaced 100mm apart. Each nail has a diameter of

5mm.

Solution:

max

3 3

7 4

5 3

4 3

5 4

10 from A to B

310 150 250 90I= 7.2 10

12 12

take the Q for the top board

Q=y A = 30 250 60 4.5 10

10 4.5 1062.5 /

7.2 10

V kN

mm

VQq

I

mm

kN mq kN m

m

Half of this shear flow comes along each edge of the board

Shear flow along one edge of board is 31.25kN/m

Shear force Vnail in a nail is shear flow along edge * spacing:

Computing Deflections in Beams:

1. Determinate Beams:

a. Starting with M(x), integrate to find (x).

b. Integrate (x) to obtain v(x).

2. Indeterminate Beams: (Use kinematic constraints from

excess support reactions to solve for the coefficients of

integration.)

a. Starting with w(x), integrate to obtain V(x)

b. Integrate V(x) to obtain M(x)

c. Integrate M(x) to obtain (x)

d. Integrate (x) to obtain v(x)

4

4

3

3

2

2

( )

( )

( )( )

( )

dV d vw x EI

dx dx

dM d vV x EI

dx dx

d M x d vx

dx EI dx

dvx

dx

2

4

31.25 / 0.1 3.125

3.125159

.005

nail

nailave

nail

V kN m m kN

V kNMPa

A m


10

Example: Compute the slope of the elastic curve at A and B, and the deflection at C. EI is

constant. Solution:

Principle of Superposition Statically

indeterminate beams are those that have more supports than there are relevant equilibrium equations. For such beams, one cannot just use the equations of static equilibrium to solve for the support reactions.

Solving for the displacements, slopes, shears, and moments in statically

1

2

2

2

3 2

3

4 3

3 4

4 3

3

3

3

( )

since V(0) = 2 2

( )2 2

since M(0)=02 2

EI (x)=6 4

24 12

since v(0)=024 12

EIv(L)=0 c24

V x wx c

wL wLwx

wx wLxM x c

wx wLx

wx wLxc

wx wLxEIv x c x c

wx wLxc x

wL

3 2 3( ) 224

wxv x x Lx L

EI

4

3

3

5

2 384

024

24

C

A

B

L wLv v

EI

wL

EI

wLL

EI

RB

v(x)=v1(x) + v2(x)

v1(x)

v2(x, RB)

Solve for RB

such that:

v1 (L) + v2(L)=0


11

indeterminate beams is actually very straightforward, because whenever a support condition exists for the beam, there is a corresponding kinematic constraint on the beam.

Usage of superposition is one way to solve statically indeterminate beam problems.

G. Euler Buckling Loads with different types of supports

Buckling is primarily a concern or issue in long slender members subjected to axial compression. The critical axial compression load that will cause buckling is computed as where: E is the Young’s modulus, I is the minimum moment of inertia of the cross-section, and L is the length of the member. Depending upon the manner in which the compression member is restrained at its ends, the effective length factor K will change as shown in the diagrams below.

2

2

KL

EIPcr


12

H. Energy Methods

1. Elastic Strain Energies

a. Axial Loading

dxAE

NAdx

EA

NAdx

EdV

EdVuU xx

2222

2

2

222

b. Bending

dxIE

xMdxdAy

EI

xMdV

EI

yxMdVuU

2

)(

2

)(

2

)( 22

2

2

2

2

c. Torsional Deformation

dxGJ

TdxdA

GJ

TAdx

GJ

TdV

GdVuU

2222

22

2

2

2

222

2. Conservation of Energy (quasi-static systems)

2

PU

Δ Knowing U and P, can solve for

Using Conservation of Energy, displacement can only be computed at the same point and in the same direction as the applied load.

3. Principle of Virtual Work (More General)

A mechanical system will be subjected to a real set of loads F, and this will result in real

strains in the material that comprises the system.

Imagine that before the system is subjected to the real loads F it is first subjected to an

infinitesimally small virtual load fv at the point in the system where we desire to know the

L

dxEI

M

PP

U

0

2

2

22


13

displacement, in the direction we wish to quantify the displacement. This virtual load

will give rise to equilibrium virtual stresses v and virtual strains v in the system.

Next, the real loads F are applied to the system resulting in real displacements

throughout the system as well as real equilibrium stresses and strains in the system.

Externally, the work done by the virtual force as the structure undergoes real

displacements is: δf vvW

Internally, the virtual strain energy associated with the real strains and the virtual stresses

is V

vv dVU εσ :

Equating the external virtual work with the internal virtual strain energy we get:

v

V

vvv UdVW εσδf :

The real displacement magnitude at the location of the virtual force and parallel to the

direction of the virtual force is:

V

v

vv

v dVεσff

δf:

1

Example:

Consider the cantilever beam shown. If we wanted to compute the vertical displacement at A, we could use conservation of energy.

But if we wanted to compute the slope at B, the principle of virtual work might be more direct.


14

First, we apply a virtual moment at B as shown below.

.

The resulting virtual moment distribution mv(x) in the beam from the virtual moment at B would be as follows:

1 0 x L/2( )

0 x L 2 vm x

The virtual bending stresses in the beam would be as follows:

0 x L/2( , )

0 x L 2

vv

ym y

x y II

When the real load P is applied to the tip of the beam, the real moment distribution in the beam is:

L

xPLxM 1)(

The resulting real bending strains in the beam are ( )

( , )M x y

x yEI

:

The virtual strain energy in the beam would be as follows:

The external virtual work is: 1v BW

Equating the external virtual work and virtual strain energy gives: 23

8B

PL

EI

22

2 2

0 0

/2 /2

0 0

2

( ) ( ) ( ) ( ) ( ) ( )

1 ( ) 1

2 8

3

8

L L

v v vv

V A

L L

m x M x y m x M x m x M xU dV y dA dx dx

EI EI EI

M x PL x PL L Ldx dx

EI EI L EI

PL

EI


15

I. STRESS TRANSFORMATIONS

1. Plane Stress

A state of plane stress at a point is

defined by specifying the normal and

shear stress components on two

perpendicular planes: For example:x,

y, xy

All combinations of normal and shear

on other planes passing through the

same point plot on a circle (Mohr’s

circle) in - space.

The radius R of the circle and the

center C on the -axis are given by the

following relations:

2

2

2xy

yxR

;

2

yx

avgC

Once the Mohrs Circle is known, the in-plane principal stresses are easily found:

o Major principal stress: 1 C R

o Minor principal stress: 3 C R

In plane maximum shear stress: max R

2. Triaxial States of Stress:

First find all three principal stresses:

1 = major principal stress

2 = intermediate principal stress

3 = minor principal stress

Then the absolute maximum shear stress is: 2

31

max

The normal stress that corresponds with the absolute max shear stress:

2

31

ave


16

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